Evaluating the Lie derivative of a tensor

Question

Let $X$ be a vector field, $L_X$ the Lie derivative and $T$ a tensor of rank $(1,2)$. I would like to find $(L_X T)^a{}_{bc}$. What I have tried so far $$\begin{align*}(L_X T)^a{}_{bc}&=(L_XT)(dx^a,\partial_b,\partial_c)\\ &= L_X(T(dx^a,\partial_b,\partial_c)) \\ &- T(L_X(dx^a),\partial_b,\partial_c)- T(dx^a,L_X(\partial_b),\partial_c)- T(dx^a,\partial_b,L_X(\partial_c))\\ &= L_X(T^a{}_{bc})- T(L_X(dx^a),\partial_b,\partial_c)- T(dx^a,L_X(\partial_b),\partial_c)- T(dx^a,\partial_b,L_X(\partial_c)) \\ &= X^kT^a{}_{bc,k}- T(L_X(dx^a),\partial_b,\partial_c)- T(dx^a,L_X(\partial_b),\partial_c)- T(dx^a,\partial_b,L_X(\partial_c)).\end{align*}$$ If I'm not mistaken we also have $L_X(\partial_k) = [X,\partial_k]= -X^m{}_{,k}\partial_m$ which then leads to $$(L_XT)^a{}_{bc}= X^kT^a{}_{bc,k} + X^m{}_{,b}T^a{}_{mc} + X^m{}_{,c}T^a{}_{bm} - T(L_X(dx^a),\partial_b,\partial_c).$$ And this is where I'm stuck. I know that $L_X$ leaves the rank of a tensor unchanged so $L_X(dx^a)$ will presumably be a linear combination of $dx^k$, but I don't really know how to find the coefficients of this linear combination.

I'd appreciate any suggestions on how one might evaluate the expression $L_X(dx^a)$.

Answer 1

You can deduce this from \begin{align*} &0 = L_X(\delta^a_{\ b}) = L_X(dx^a(\partial_b)) = L_X(dx^a)(\partial_b)+dx^a(L_X(\partial_b)) \\ \implies & L_X(dx^a)(\partial_b) = -dx^a(L_X(\partial_b)) = X^a_{\ \ ,b}. \end{align*} So $L_X(dx^a) = X^a_{\ \ ,b}dx^b$.

Alternatively, if you know the fact that $L_X$ commutes with $d$, you can simply say $$ L_X(dx^a) = d(L_Xx^a) = d(Xx^a) = d(X^a) = X^a_{\ \ ,b}dx^b. $$

Answer 2

Inspired by the answer of @user17945. I just noticed that we proved Cartans formula in the lecture. So if $\iota_X$ is the interior product, we have $$L_X(dx^a)= (d\circ \iota_X + \iota_X \circ d)(dx^a) \overset{d^2=0}{=} d(\iota_X (dx^a)) = d(Xx^a) = d(X^a)= X^a{}_{,b}dx^b.$$